Answer
$a_1=\sqrt 2\\ a_2=\sqrt{2+a_{1}}=\sqrt{2+\sqrt2} \\ a_3=\sqrt{2+a_{2}}=\sqrt{2+\sqrt{2+\sqrt2}} \\ a_4=\sqrt{2+a_{3}}=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}} \\ a_5=\sqrt{2+a_{4}}=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}$
Work Step by Step
We are given that $a_n=\sqrt{2+a_{n-1}}$ and $a_1=\sqrt2$. In order to determine the remaining values, we will have to substitute $n=2,3,4,5$ into the given sequence. We use previous terms to compute the successive terms, as follows:
$a_2=\sqrt{2+a_{1}}=\sqrt{2+\sqrt2} \\ a_3=\sqrt{2+a_{2}}=\sqrt{2+\sqrt{2+\sqrt2}} \\ a_4=\sqrt{2+a_{3}}=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}} \\ a_5=\sqrt{2+a_{4}}=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt2}}}}$
